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$$\frac{1}{2} \log(x+2)=2$$

I'm decently good at logarithms but this one seems to be tricky, when I did it myself I got a negative decimal as my answer but I'm not 100% confident in it, and I would really appreciate some help!

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Is your equation $(1/2)\log(x+2)=2$ or $1/(2\log(x+2))=2$? –  David Mitra Jan 12 '13 at 19:10
    
@user57055: Welcome to MSE! Please format questions using MathJax as it makes readability much easier. Also, if this is homework, please tag it as such (and it is okay if it is). It is helpful to the MSE community for you to post what you did so they can identify where you may have gone astray. Regards –  Amzoti Jan 12 '13 at 19:10
    
Obtaining a negative value $x$ as a solution is fine, here, for the equation ${1\over 2\log(x+2)}=2$ (where $\log$ is the natural logarithm). Solving this, we obtain the solution $x=e^{1/4}-2$. This is negative; and no doubt you know you can't take the logarithm of a negative number. But we don't do that here: we take the logarithm of $x\color{maroon}{+2}$, which is positive for our $x$. And this solution does "work". It's always a good idea to check your solutions when solving equations. –  David Mitra Jan 12 '13 at 19:16
    
The equation is (1/2)log(x+2)=2 David –  user57055 Jan 12 '13 at 19:20
    
You shouldn't have obtained a negative solution then. (The solution is $e^4-2$ for the natural logarithm, and $10^4-2$ for the common logarithm.) –  David Mitra Jan 12 '13 at 19:25

2 Answers 2

You have

$\frac{1}{2} \log(x+2)=2$

multiply both sides for 2

$\log(x+2)=4$

Now, I suppose the logarithm base is $e$ so, raise $e$ to both sides of the equation

$(x+2)=e^4$

so, $x=e^4-2$.

Similarly, if the base of the logarithm is 10, the answer is $x=10^4-2$

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Thank you so much, finally put all the pieces together and found out what I did wrong. The log base of 10 is already built into the word log on my TI-84+ so without the base of 10, the problem would be wrong. I had to pull the base of 10 out mechanically and actually put it into the equation. –  user57055 Jan 12 '13 at 19:55
    
@user57055 : Mathematicians usually take $\log x$ to mean $\log_e x$. Some others often take it to mean $\log_{10} x$. –  Michael Hardy Jan 12 '13 at 20:32

what are you guys talking about? $\log_{e}$ is written as $\ln$. The easiest way to solve this problem is simply writing logarithm in exponent form. First, move the $1/2$ up to an exponent (law of logs). Now we have $\log(x+2)^{1/2} = 2$. From there, (and it is base ten), rewrite in exponent form --> the base, raised to the answer, equals whats in from of the log. So $$10^2 = (x+2)^{1/2}$$ $$100 = \sqrt{x+2}$$ $$10000 = x+2$$ $$99998 = x$$

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See Michael Hardy's comment to the other answer, and note that the other answer covers both bases. Also, the last line has one too many 9. –  epimorphic Dec 8 at 5:02

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